求1~n与x互质的数的个数（6个题、容斥原理）

HDU 4135、POJ 2773、HDU 1695、HDU 2841、ZOJ 2836、HDU 1796

HDU 4135 Co-prime

题意: 求[l,r]与x互质的数的个数

分析: 裸题

代码:

//
//  Created by TaoSama on 2015-10-24
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

typedef long long LL;

vector<int> factor;
void fact(int x) {
factor.clear();
for(int i = 2; i * i <= x; ++i) {
if(x % i == 0) {
factor.push_back(i);
while(x % i == 0) x /= i;
}
if(x == 1) break;
}
if(x > 1) factor.push_back(x);
}

LL calc(LL x) {
LL ret = 0; //not co-prime
int sz = factor.size();
for(int s = 1; s < 1 << sz; ++s) {
int cnt = 0, mul = 1;
for(int i = 0; i < sz; ++i)
if(s >> i & 1) ++cnt, mul *= factor[i];
if(cnt & 1) ret += x / mul;
else ret -= x / mul;
}
return x - ret;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
int kase = 0;
while(t--) {
LL l, r, k; scanf("%I64d%I64d%I64d", &l, &r, &k);
fact(k);
printf("Case #%d: %I64d\n", ++kase, calc(r) - calc(l - 1));
}
return 0;
}

POJ 2773 Happy 2006

题意: 求第k个[l,r]与x互质的数

分析: 裸题上二分一下

代码:

//
//  Created by TaoSama on 2015-10-24
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, k;

typedef long long LL;

vector<int> factor;
void fact(int x) {
factor.clear();
for(int i = 2; i * i <= x; ++i) {
if(x % i == 0) {
factor.push_back(i);
while(x % i == 0) x /= i;
}
if(x == 1) break;
}
if(x > 1) factor.push_back(x);
}

LL calc(LL x) {
LL ret = 0;
int sz = factor.size();
for(int s = 1; s < 1 << sz; ++s) {
int cnt = 0, mul = 1;
for(int i = 0; i < sz; ++i)
if(s >> i & 1) ++cnt, mul *= factor[i];
if(cnt & 1) ret += x / mul;
else ret -= x / mul;
}
return x - ret;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d%d", &n, &k) == 2) {
fact(n);
LL l = 1, r = 1e10;
while(l <= r) {
LL m = l + r >> 1;
if(calc(m) < k) l = m + 1;
else r = m - 1;
}
printf("%I64d\n", l);
}
return 0;
}

HDU 1695 GCD

题意: 求[1,l]和[1,r]两个区间元素对gcd(x,y)=k的个数

分析: 除k后变成gcd(x,y)=1,然后枚举x,保证x<y就好了,特判k=0

代码:

//
//  Created by TaoSama on 2015-10-24
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

vector<int> factor
;
void gao() {
bool vis
= {};
for(int i = 2; i < N; ++i) {
if(vis[i]) continue;
for(int j = i; j < N; j += i) {
vis[j] = true;
factor[j].push_back(i);
}
}
}

typedef long long LL;

int calc(int k, int x) {
int ret = 0;
int sz = factor[k].size();
for(int s = 1; s < 1 << sz; ++s) {
int cnt = 0, mul = 1;
for(int i = 0; i < sz; ++i) {
if(s >> i & 1) ++cnt, mul *= factor[k][i];
}
if(cnt & 1) ret += x / mul;
else ret -= x / mul;
}
return x - ret;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
int kase = 0;
gao();
while(t--) {
int l, r, k; scanf("%d%d%d%d%d", &l, &l, &r, &r, &k);
if(k == 0) {printf("Case %d: 0\n", ++kase); continue;}
l /= k, r /= k;
if(l > r) swap(l, r);
LL ans = 0;
for(int i = 1; i <= l; ++i)
ans += calc(i, r) - calc(i, i - 1);
printf("Case %d: %I64d\n", ++kase, ans);
}
return 0;
}

HDU 2841 Visible Trees

题意: 站在(0,0),看(1,1)到(n,m)这个网格的格点,能看到多格点

分析: 易知只有gcd(x,y)=1才能被看到,枚举小的那个坐标,没有然后了

代码:

//
//  Created by TaoSama on 2015-10-24
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

vector<int> factor
;
void gao() {
bool vis
= {};
for(int i = 2; i < N; ++i) {
if(vis[i]) continue;
for(int j = i; j < N; j += i) {
vis[j] = true;
factor[j].push_back(i);
}
}
}

int calc(int k, int x) {
int ret = 0;
int sz = factor[k].size();
for(int s = 1; s < 1 << sz; ++s) {
int cnt = 0, mul = 1;
for(int i = 0; i < sz; ++i)
if(s >> i & 1) ++cnt, mul *= factor[k][i];
if(cnt & 1) ret += x / mul;
else ret -= x / mul;
}
return x - ret;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
gao();
while(t--) {
int n, m; scanf("%d%d", &n, &m);
if(n > m) swap(n, m);
long long ans = 0;
for(int i = 1; i <= n; ++i) ans += calc(i, m);
printf("%I64d\n", ans);
}
return 0;
}

ZOJ 2836 Number Puzzle

题意: 求[1,m]中能被列表中至少一个数整除 的数的个数

分析: 同理,不过乘的时候要求lcm,因为可能不互质

代码:

//
//  Created by TaoSama on 2015-10-24
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, a[15];

typedef long long LL;

LL calc() {
LL ret = 0;
for(int s = 1; s < 1 << n; ++s) {
int cnt = 0, mul = 1;
for(int i = 0; i < n; ++i)
if(s >> i & 1) ++cnt, mul = mul / __gcd(mul, a[i]) * a[i];
if(cnt & 1) ret += m / mul;
else ret -= m / mul;
}
return ret;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d%d", &n, &m) == 2) {
for(int i = 0; i < n; ++i) scanf("%d", a + i);
printf("%lld\n", calc());
}
return 0;
}

HDU 1796 How many integers can you find

题意: 跟上一个题题意一样

分析: [1,n)右边是开区间,−−n,然后列表数字≥0,先去0,然后就一样了

代码:

//
//  Created by TaoSama on 2015-10-24
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, a[15];

typedef long long LL;

LL calc() {
LL ret = 0;
vector<int> factor;
for(int i = 0; i < n; ++i) if(a[i]) factor.push_back(a[i]);
n = factor.size();
for(int s = 1; s < 1 << n; ++s) {
int cnt = 0, mul = 1;
for(int i = 0; i < n; ++i)
if(s >> i & 1) ++cnt, mul = mul / __gcd(mul, a[i]) * a[i];
if(cnt & 1) ret += m / mul;
else ret -= m / mul;
}
return ret;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d%d", &m, &n) == 2) {
--m;
for(int i = 0; i < n; ++i) scanf("%d", a + i);
printf("%I64d\n", calc());
}
return 0;
}